Cho hàm số \(y=f\left( x \right)=2019\ln \left( {{e}^{\frac{x}{2019}}}+\sqrt{e} \right)\). Tính giá trị biểu thức \(A=f'\left( 1 \right)+f'\left( 2 \right)+...+f'\left( 2018 \right...

* Hướng dẫn giải

Ta có \(y'=f'\left( x \right)=2019\frac{{{\left( {{e}^{\frac{x}{2019}}}+\sqrt{e} \right)}^{'}}}{\left( {{e}^{\frac{x}{2019}}}+\sqrt{e} \right)}=\frac{{{e}^{\frac{x}{2019}}}}{{{e}^{\frac{x}{2019}}}+\sqrt{e}}\)

Do đó

\(f'\left( x \right)+f'\left( 2019-x \right)=\frac{{{e}^{\frac{x}{2019}}}}{{{e}^{\frac{x}{2019}}}+\sqrt{e}}+\frac{{{e}^{\frac{2019-x}{2019}}}}{{{e}^{\frac{2019-x}{2019}}}+\sqrt{e}}=\frac{{{e}^{\frac{x}{2019}}}}{{{e}^{\frac{x}{2019}}}+\sqrt{e}}+\frac{{{e}^{1-\frac{x}{2019}}}}{{{e}^{1-\frac{x}{2019}}}+\sqrt{e}}\)

\(=\frac{{{e}^{\frac{x}{2019}}}}{{{e}^{\frac{x}{2019}}}+\sqrt{e}}+\frac{e}{e+\sqrt{e}.{{e}^{\frac{x}{2019}}}}=\frac{{{e}^{\frac{x}{2019}}}}{{{e}^{\frac{x}{2019}}}+\sqrt{e}}+\frac{\sqrt{e}}{\sqrt{e}+{{e}^{\frac{x}{2019}}}}=1\)

Bởi vậy \(2A=\left[ f'\left( 1 \right)+f'\left( 2018 \right) \right]+\left[ f'\left( 2 \right)+f'\left( 2017 \right) \right]+...+\left[ f'\left( 2018 \right)+f'\left( 1 \right) \right]=2018\)

Nên \(A=\frac{2018}{2}=1009\)